Rate of precession

The rate of precession depends on the inclination of the orbital plane to the equatorial plane, as well as the orbital eccentricity.

For a satellite in a prograde orbit around Earth, the precession is westward (nodal regression), that is, the node and satellite move in opposite directions.^[1] A good approximation of the precession rate is

${\displaystyle \omega _{\mathrm {p} }=-{\frac {3}{2}}{\frac {{R_{\mathrm {E} }}^{2}}{\left(a\left(1-e^{2}\right)\right)^{2}}}J_{2}\omega \cos i}$

where

ω_p is the precession rate (in rad/s),

R_E is the body's equatorial radius (6378137 m for Earth),

a is the semi-major axis of the satellite's orbit,

e is the eccentricity of the satellite's orbit,

ω is the angular velocity of the satellite's motion (2π radians divided by its period in seconds),

i is its inclination,

J₂ is the body's "second dynamic form factor" ^[2](−√5C₂₀^[3] = 1.08262668×10⁻³ for Earth).

This last quantity is related to the oblateness as follows:

${\displaystyle J_{2}={\frac {2\varepsilon _{\mathrm {E} }}{3}}-{\frac {{R_{\mathrm {E} }}^{3}{\omega _{\mathrm {E} }}^{2}}{3GM_{\mathrm {E} }}}}$

where

ε_E is the central body's oblateness,

R_E is central body's equatorial radius (6378137 m for Earth),

ω_E is the central body's rotation rate (7.292115×10⁻⁵ rad/s for Earth),

GM_E is the product of the universal constant of gravitation and the central body's mass (3.986004418×10¹⁴ m³/s² for Earth).

The nodal progression of low Earth orbits is typically a few degrees per day to the west (negative). For a satellite in a circular (e = 0) 800 km altitude orbit at 56° inclination about Earth:

${\displaystyle {\begin{aligned}R_{\mathrm {E} }&=6.378\,137\times 10^{6}{\text{ m}}\\J_{2}&=1.082\,626\,68\times 10^{-3}\end{aligned}}}$

The orbital period is 6052.4 s, so the angular velocity is 0.001038 rad/s. The precession is therefore

${\displaystyle {\begin{aligned}\omega _{\mathrm {p} }&=-{\frac {3}{2}}\cdot {\frac {6\,378\,137^{2}}{\left(7\,178\,137\left(1-0^{2}\right)\right)^{2}}}\cdot \left(1.082\,626\,68\times 10^{-3}\right)\cdot 0.001\,038\cdot \cos 56^{\circ }\\&=-7.44\times 10^{-7}{\text{ rad/s}}\end{aligned}}}$

This is equivalent to −3.683° per day, so the orbit plane will make one complete turn (in inertial space) in 98 days.

The apparent motion of the sun is approximately +1° per day (360° per year / 365.2422 days per tropical year ≈ 0.9856473° per day), so apparent motion of the sun relative to the orbit plane is about 2.8° per day, resulting in a complete cycle in about 127 days. For retrograde orbits ω is negative, so the precession becomes positive. (Alternatively, ω can be thought of as positive but the inclination is greater than 90°, so the cosine of the inclination is negative.) In this case it is possible to make the precession approximately match the apparent motion of the sun, resulting in a heliosynchronous orbit.

The ${\displaystyle J_{2}}$ used in this equation is the dimensionless coefficient ${\displaystyle {\tilde {J_{2}}}=-{\frac {J_{2}}{GM_{E}R_{E}^{2}}}}$ from the geopotential model or gravity field model for the body.